hw1sol - Instructor Christopher Wayne Walker Ph.D TA...

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Unformatted text preview: Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 EE 132A Homework 1 Solutions Problem 1. For a negationslash = 0 prove the scaling property of the Fourier transform: x ( at ) ←→ 1 | a | X ( f/a ) . Solution: From the definition of the Fourier transform: F [ x ( at )] = integraldisplay ∞-∞ x ( at ) e- j 2 πft dt Using the change of variables u = at , if a is positive, we obtain: F [ x ( at )] = 1 a integraldisplay ∞-∞ x ( u ) e- j 2 πfu/a du = 1 a X ( f/a ) If a is negative, we get: F [ x ( at )] = − 1 a integraldisplay ∞-∞ x ( u ) e- j 2 πfu/a du = − 1 a X ( f/a ) From these two results we conclude that F [ x ( at )] = 1 | a | X ( f/a ) Problem 2. Prove the time-shift property of the Fourier transform: x ( t − t ) ←→ e- j 2 πft X ( f ) . Solution: From the definition of the Fourier transform: F [ x ( t − t )] = integraldisplay ∞-∞ x ( t − t ) e- j 2 πft dt Using the change of variables u = t − t we obtain: F [ x ( t − t )] = integraldisplay ∞-∞ x ( u ) e- j 2 πf ( u + t ) dt = e- j 2 πft integraldisplay ∞-∞ x ( u ) e- j 2 πfu du = e- j 2 πft X ( f ) Problem 3. 1 Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 a. Compute the Fourier transform of the rectangular pulse, x ( t ) = producttext ( t ), and sketch a plot of its graph in the frequency domain. You should produce two plots (one for the magnitude of the Fourier transform and one for the phase). In the magnitude plot, clearly identify where the maximum height occurs and the value of the maximum height. Also, identify where the zero-crossings occur. b. Repeat part a for x ( t ) = producttext ( t ) cos(2 πf t ) where f = 1 kHz. Solution: a) From the definition of the Fourier transform: F [Π( t )] = integraldisplay ∞-∞ Π( t ) e- j 2 πft dt = integraldisplay 1 / 2- 1 / 2 e- j 2 πft dt When f = 0, it is evident that the result of the above integration becomes 1. When f negationslash = 0, we obtain: F [Π( t )] = e- jπf − e jπf − j 2 πf Using Euler’s formula sin θ = e jθ- e- jθ j 2 we arrive at: F [Π( t )] = braceleftBigg sin( πf ) πf ,f negationslash = 0 1 ,f = 0 bracerightBigg = sinc( f ) The magnitude and phase plots are shown in Figure 1.-5-4-3-2-1 1 2 3 4 5 0.2 0.4 0.6 0.8 1 Magnitude-5-4-3-2-1 1 2 3 4 5 1 2 3 4 Phase (rad) Figure 1: Figure of problem 3, part a b) Using Euler’s rule we can decompose the cosine as follows x ( t ) = Π( t ) e j 2 πf t + e- j 2 πf t 2 = 1 2 Π( t ) e j 2 πf t + 1 2 Π( t ) e- j 2 πf t 2 Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 From the modulation property of the Fourier transform and using linearity, we obtain X ( f ) = 1 2 F [Π( t )] vextendsingle vextendsingle vextendsingle f- f + 1 2 F [Π( t )] vextendsingle vextendsingle vextendsingle f + f = 1 2 sinc( f − f ) + 1 2 sinc( f + f ) The magnitude and phase plots are shown in Figure 2....
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